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The statement of the theorem is: If $f=0$ almost everywhere, then $\int f =0$. My question is in the proof in Folland (provided below) it seems that we are using some statement like: If $\phi$ is simple, then $\phi=0$ almost everywhere implies $\int \phi=0$ which is something we are trying to prove. I don't think I'm understanding completely the logic to why this is true. Any help clearing this up would be very helpful.

Take any $f \in L^+$, the set of all positive Lebesgue measurable functions.

If $f=0$ a.e. and $\phi$ is simple such that $0\leq\phi\leq f$, then $\phi =0$ a.e. and so $\int f=\sup_{\phi\leq f}\int \phi=0$.

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Consider first the case of simple functions: let $\phi = \sum_{i=1}^{n} a_i \chi_{E_i}$ be a simple function where the sets $E_i$'s are $\mu$-measurable. If $\phi$ is zero $\mu$-almost everywhere, then $a_i \not= 0$ iff $\mu(E_i) = 0$. Therefore, $\int \phi d\mu = \sum_{i=1}^n a_i \mu(E_i) = 0$, where the sum is zero since in every term, either $a_i = 0$ or $\mu(E_i) = 0$.

For the more general case of a positive measurable function $f$, we can just approximate $f$ by simple functions, as you noted above.

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  • $\begingroup$ Yes, I understand that part of the proof. My question is why does this proof work for the more general case. I'm not seeing why $\phi$ being simple and 0 almost everywhere imply that it's integral is 0 since this is something we are trying to prove in the first place. $\endgroup$ – user23793 Dec 22 '14 at 0:07
  • $\begingroup$ Perhaps I misunderstood, but above I show by an explicit computation that the integral of an a.e.-zero simple function is zero. Then, since the Lebesgue integral of a positive measurable function $f$ is defined to be the supremum over positive simple functions smaller than $f$, $\int f = \sup 0 = 0$. If that still does not answer your question, could you be more precise? $\endgroup$ – msteve Dec 22 '14 at 0:13
  • $\begingroup$ I'm a fool. Thank you so much. $\endgroup$ – user23793 Dec 22 '14 at 0:14

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